1.08k Separable differential equations: dy/dx = f(x)g(y)

The points \((x, y)\) on the curve \(C\) satisfy \((x + 1)(x + 2) \frac{dy}{dx} = xy\). The line with equation \(y = 2x + 5\) is the tangent to \(C\) at a point \(P\).

1. Find the coordinates of \(P\). [4]
2. Find the equation of \(C\), giving your answer in the form \(y = f(x)\). [8]

The rate of change of a certain population \(P\) at time \(t\) is modelled by the equation \(\frac{dP}{dt} = (100 - P)\). Initially \(P = 2000\).

1. Determine an expression for \(P\) in terms of \(t\). [7]
2. Describe how the population changes over time. [2]

A curve \(C\) in the \(x\)-\(y\) plane has the property that the gradient of the tangent at the point \(P(x, y)\) is three times the gradient of the line joining the point \((3, 2)\) to \(P\).

1. Express this property in the form of a differential equation. [2]

It is given that \(C\) passes through the point \((4, 3)\) and that \(x > 3\) and \(y > 2\) at all points on \(C\).

1. Determine the equation of \(C\) giving your answer in the form \(y = f(x)\). [4]

The curve \(C\) may be obtained by a transformation of part of the curve \(y = x^3\).

1. Describe fully this transformation. [2]

A car \(C\) is moving horizontally in a straight line with velocity \(v \text{ms}^{-1}\) at time \(t\) seconds, where \(v > 0\) and \(t \geq 0\). The acceleration, \(a \text{ms}^{-2}\), of \(C\) is modelled by the equation $$a = v\left(\frac{8t}{7 + 4t^2} - \frac{1}{2}\right).$$

1. In this question you must show detailed reasoning. Find the times when the acceleration of \(C\) is zero. [3] At \(t = 0\) the velocity of \(C\) is \(17.5 \text{ms}^{-1}\) and at \(t = T\) the velocity of \(C\) is \(5 \text{ms}^{-1}\).
2. By setting up and solving a differential equation, show that \(T\) satisfies the equation $$T = 2 \ln\left(\frac{7 + 4T^2}{2}\right).$$ [6]
3. Use an iterative formula, based on the equation in part (b), to find the value of \(T\), giving your answer correct to 4 significant figures. Use an initial value of 11.25 and show the result of each step of the iteration process. [2]
4. The diagram below shows the velocity-time graph for the motion of \(C\). \includegraphics{figure_7d} Find the time taken for \(C\) to decelerate from travelling at its maximum speed until it is travelling at \(5 \text{ms}^{-1}\). [1]

A gardener stores rainwater in a cylindrical container. The container has a height of 130 centimetres. The gardener empties the water from the container through a hose. The hose is attached 5 centimetres from the bottom of the container. At time \(t\) minutes after the hose is switched on, the depth of water, \(h\) centimetres, in the container decreases at a rate which is proportional to \(h - 5\) Initially the container of water is full, and the depth of water is decreasing at a rate of 1.5 centimetres per minute.

1. Show that $$\frac{dh}{dt} = -0.012(h - 5)$$ [3 marks]
2. Solve the differential equation $$\frac{dh}{dt} = -0.012(h - 5)$$ to find an expression for \(h\) in terms of \(t\) [5 marks]
3. Find the time taken for the container to be half empty. Give your answer to the nearest minute. [2 marks]

The height \(x\) metres, of a column of water in a fountain display satisfies the differential equation \(\frac{dx}{dt} = \frac{8\sin 2t}{3\sqrt{x}}\), where \(t\) is the time in seconds after the display begins.

1. Solve the differential equation, given that initially the column of water has zero height. Express your answer in the form \(x = f(t)\) [7 marks]
2. Find the maximum height of the column of water, giving your answer to the nearest cm. [1 mark]

A market trader notices that daily sales are dependent on two variables: number of hours, \(t\), after the stall opens total sales, \(x\), in pounds since the stall opened. The trader models the rate of sales as directly proportional to \(\frac{8 - t}{x}\) After two hours the rate of sales is £72 per hour and total sales are £336

1. Show that $$x \frac{dx}{dt} = 4032(8 - t)$$ [3 marks]
2. Hence, show that $$x^2 = 4032t(16 - t)$$ [3 marks]
3. The stall opens at 09.30.
   1. The trader closes the stall when the rate of sales falls below £24 per hour. Using the results in parts (a) and (b), calculate the earliest time that the trader closes the stall. [6 marks]
   2. Explain why the model used by the trader is not valid at 09.30. [2 marks]

Solve the differential equation $$\frac{dt}{dx} = \frac{\ln x}{x^2 t} \quad \text{for } x > 0$$ given \(x = 1\) when \(t = 2\) Write your answer in the form \(t^2 = f(x)\) [7 marks]

A particle moves so that its acceleration, \(a\text{ ms}^{-2}\), at time \(t\) seconds may be modelled in terms of its velocity, \(v\text{ ms}^{-1}\), as $$a = -0.1v^2$$ The initial velocity of the particle is \(4\text{ ms}^{-1}\)

1. By first forming a suitable differential equation, show that $$v = \frac{20}{5 + 2t}$$ [6 marks]
2. Find the acceleration of the particle when \(t = 5.5\) [2 marks]

A circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface. As the weed grows, it covers an area of \(A\) square metres. A simple model assumes that the weed grows so that the rate of increase of its area is proportional to \(A\).

1. Show that the area covered by the weed can be modelled by $$A = Be^{kt}$$ where \(B\) and \(k\) are constants and \(t\) is time in days since the weed was first noticed. [4 marks]
2. When it was first noticed, the weed covered an area of 0.25 m². Twenty days later the weed covered an area of 0.5 m²
   1. State the value of \(B\). [1 mark]
   2. Show that the model for the area covered by the weed can be written as $$A = 2^{\frac{t}{20} - 2}$$ [4 marks]
   3. How many days does it take for the weed to cover half of the surface of the pond? [2 marks]
3. State one limitation of the model. [1 mark]
4. Suggest one refinement that could be made to improve the model. [1 mark]

In a chemical reaction, the mass \(m\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dm}{dt} = \frac{m}{t(1 + 2t)}.$$ At time 1 minute, the mass of the chemical is 1 gram.

1. Solve the differential equation to show that \(m = \frac{3t}{(1 + 2t)}\). [8]
2. Hence
   1. find the time when the mass is 1.25 grams, [2]
   2. show what happens to the mass of the chemical as \(t\) becomes large. [2]

A small, hollow, plastic ball, of mass \(m\) kg is at rest at a point \(O\) on a polished horizontal surface. The ball is attached to two identical springs. The other ends of the springs are attached to the points \(P\) and \(Q\) which are 1.8 metres apart on a straight line through \(O\). The ball is struck so that it moves away from \(O\), towards \(P\) with a speed of 0.75 m s\(^{-1}\). As the ball moves, its displacement from \(O\) is \(x\) metres at time \(t\) seconds after the motion starts. The force that each of the springs applies to the ball is \(12.5mx\) newtons towards \(O\). The ball is to be modelled as a particle. The surface is assumed to be smooth and it is assumed that the forces applied to the ball by the springs are the only horizontal forces acting on the ball.

1. Find the minimum distance of the ball from \(P\), in the subsequent motion. [5 marks]

A particle \(P\) of mass \(2\) kg moves along the \(x\)-axis. At time \(t = 0\), \(P\) passes through the origin \(O\) with speed \(3\) m s\(^{-1}\). At time \(t\) seconds the displacement of \(P\) from \(O\) is \(x\) m and the velocity of \(P\) is \(v\) m s\(^{-1}\), where \(t \geqslant 0\), \(x \geqslant 0\) and \(v \geqslant 0\). While \(P\) is in motion the only force acting on \(P\) is a resistive force \(F\) of magnitude \((v^2 + 1)\) N acting in the negative \(x\)-direction.

1. Find an expression for \(v\) in terms of \(x\). [5]
2. Determine the distance travelled by \(P\) while its speed drops from \(3\) m s\(^{-1}\) to \(2\) m s\(^{-1}\). [2]

Particle \(Q\) is identical to particle \(P\). At a different time, \(Q\) is moving along the \(x\)-axis under the influence of a single constant resistive force of magnitude \(1\) N. When \(t' = 0\), \(Q\) is at the origin and its speed is \(3\) m s\(^{-1}\).

1. By comparing the motion of \(P\) with the motion of \(Q\) explain why \(P\) must come to rest at some finite time when \(t < 6\) with \(x < 9\). [3]
2. Sketch the velocity-time graph for \(P\). You do not need to indicate any values on your sketch. [1]
3. Determine the maximum displacement of \(P\) from \(O\) during \(P\)'s motion. [2]

A particle P, of mass \(m\), moves on a rough horizontal table. P is attracted towards a fixed point O on the table by a force of magnitude \(\frac{kmg}{x^2}\), where \(x\) is the distance OP. The coefficient of friction between P and the table is \(\mu\). P is initially projected in a direction directly away from O. The velocity of P is first zero at a point A which is a distance \(a\) from O.

1. Show that the velocity \(v\) of P, when P is moving away from O, satisfies the differential equation $$\frac{\mathrm{d}}{\mathrm{d}x}(v^2) + \frac{2kg}{x^2} + 2\mu g = 0.$$ [3]
2. Verify that $$v^2 = 2gk\left(\frac{1}{x} - \frac{1}{a}\right) + 2\mu g(a-x).$$ [3]
3. Find, in terms of \(k\) and \(a\), the range of values of \(\mu\) for which P remains at A. [2]

The variable \(y\) satisfies the differential equation $$2\frac{dy}{dx} = 5 - 2y, \quad \text{where } x \geqslant 0.$$ Given that \(y = 1\) when \(x = 0\), find an expression for \(y\) in terms of \(x\). [5]

The rate of change of a variable \(y\) with respect to \(x\) is directly proportional to \(y\).

1. Write down a differential equation satisfied by \(y\). [1]
2. When \(x = 1\) and \(y = 0.5\), the rate of change of \(y\) with respect to \(x\) is 2. Find \(y\) when \(x = 3\). [6]

An object of mass \(0 \cdot 5\) kg is thrown vertically upwards with initial speed \(24\) ms\(^{-1}\). The velocity of the object at time \(t\) seconds is \(v\) ms\(^{-1}\). During the upward motion, the object experiences a resistance to motion \(RN\), where \(R\) is proportional to \(v\). When the velocity of the object is \(0 \cdot 2\) ms\(^{-1}\) the resistance to motion is \(0 \cdot 08\) N.

1. Show that the upward motion of the object satisfies the differential equation $$\frac{\mathrm{d}v}{\mathrm{d}t} = -9 \cdot 8 - 0 \cdot 8\,v.$$ [3]
2. Find an expression for \(v\) at time \(t\). [6]
3. Determine the value of \(t\) when the object is at the highest point of the motion. [2]

The height \(x\) metres, of a column of water in a fountain display satisfies the differential equation \(\frac{dx}{dt} = -\frac{8\sin 2t}{3\sqrt{x}}\), where \(t\) is the time in seconds after the display begins. Solve the differential equation, given that initially the column of water has zero height. Express your answer in the form \(x = f(t)\) [7 marks]

Find the general solution of the differential equation $$(2x^3 - 3x^2 - 11x + 6)\frac{dy}{dx} = y(20x - 35).$$ Give your answer in the form \(y = f(x)\). [9]

Solve the differential equation $$2\cot x \frac{dy}{dx} = (4 - y^2)$$ for which \(y = 0\) at \(x = \frac{\pi}{3}\), giving your answer in the form \(\sec^2 x = g(y)\). [8]

A population of meerkats is being studied. The population is modelled by the differential equation $$\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{1}{22}P(11 - 2P), \quad t \geq 0, \quad 0 < P < 5.5$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that there were 1000 meerkats in the population when the study began, determine the time taken, in years, for this population of meerkats to double. [7]